Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $k = \dfrac{20r^2 - 15r}{4} \times \dfrac{-6}{36r - 27} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $k = \dfrac{ (20r^2 - 15r) \times -6 } { 4 \times (36r - 27) } $ $ k = \dfrac {-6 \times 5r(4r - 3)} {4 \times 9(4r - 3)} $ $ k = \dfrac{-30r(4r - 3)}{36(4r - 3)} $ We can cancel the $4r - 3$ so long as $4r - 3 \neq 0$ Therefore $r \neq \dfrac{3}{4}$ $k = \dfrac{-30r \cancel{(4r - 3})}{36 \cancel{(4r - 3)}} = -\dfrac{30r}{36} = -\dfrac{5r}{6} $